As a matter of fact your sum diverges, a little manipulation shows that
$$\sum_{n \leq X} \frac{(-1)^n}{\phi(n)} = \sum_{n \leq X, 2|n} \frac{1}{\phi(n)} -
\sum_{n \leq X, (n,2)=1} \frac{1}{\phi(n)} = \sum_{n \leq X/2} \frac{1}{\phi(2n)}
-\sum_{n \leq X,(n,2)=1}\frac{1}{\phi(n)}$$
The above equala to
$$\sum_{n \leq X/2, (n,2)=1} \frac{1}{\phi(2n)} + \sum_{n \leq X/2, 2|n} \frac{1}{\phi(2n)}
- \sum_{n \leq X, (n,2)=1} \frac{1}{\phi(n)}$$ By multiplicativity of $\phi(n)$ we have $\phi(2n) = \phi(n)$ when $(n,2)=1$. Thus the first sum above is a sum over $1/\phi(n)$ and the above equation simplifies to
$$ - \sum_{X/2 < n \leq X, (n,2)=1} \frac{1}{\phi(n)} + \sum_{n \leq X/4} \frac{1}{\phi(4n)}$$ It follows that
$$\sum_{n \leq X} \frac{(-1)^n}{\phi(n)}= -\sum_{X/2 < n \leq X, (n,2)=1} \frac{1}{\phi(n)} + \sum_{n \leq X/4} \frac{1}{\phi(4n)} \sim c \cdot \log{X}$$
because the first sum on the right converges to a constant, while the second sum on the right is asymptotically $c \cdot \log{X}$.

EDIT: Put details, erased mention of an earlier confusion about $(-1)^{n+1}$ not being a multiplicative function :P (it is!)